3.9.31 \(\int \frac {-1280+960 x-240 x^2+20 x^3-20 e^{3 e^3} x^3+e^{2 e^3} (-240 x^2+60 x^3)+(10 x+60 x^2+75 x^3-10 x^4) \log ^2(4)+e^{e^3} (-960 x+480 x^2-60 x^3+(5 x^3+10 x^4) \log ^2(4))}{256-192 x+48 x^2-4 x^3+4 e^{3 e^3} x^3+e^{2 e^3} (48 x^2-12 x^3)+e^{e^3} (192 x-96 x^2+12 x^3)} \, dx\)

Optimal. Leaf size=37 \[ -4+5 \left (-x+\frac {(1+2 x)^2 \log ^2(4)}{16 \left (-1+e^{e^3}+\frac {4}{x}\right )^2}\right ) \]

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Rubi [B]  time = 0.58, antiderivative size = 177, normalized size of antiderivative = 4.78, number of steps used = 4, number of rules used = 2, integrand size = 175, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6, 2074} \begin {gather*} \frac {5 x^2 \log ^2(4)}{4 \left (1-e^{e^3}\right )^2}-\frac {5 x \left (4+12 e^{2 e^3}-4 e^{3 e^3}-9 \log ^2(4)-e^{e^3} \left (12-\log ^2(4)\right )\right )}{4 \left (1-e^{e^3}\right )^3}-\frac {5 \left (9-e^{e^3}\right ) \left (17-e^{e^3}\right ) \log ^2(4)}{2 \left (1-e^{e^3}\right )^4 \left (4-\left (1-e^{e^3}\right ) x\right )}+\frac {5 \left (9-e^{e^3}\right )^2 \log ^2(4)}{\left (1-e^{e^3}\right )^4 \left (4-\left (1-e^{e^3}\right ) x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1280 + 960*x - 240*x^2 + 20*x^3 - 20*E^(3*E^3)*x^3 + E^(2*E^3)*(-240*x^2 + 60*x^3) + (10*x + 60*x^2 + 75
*x^3 - 10*x^4)*Log[4]^2 + E^E^3*(-960*x + 480*x^2 - 60*x^3 + (5*x^3 + 10*x^4)*Log[4]^2))/(256 - 192*x + 48*x^2
- 4*x^3 + 4*E^(3*E^3)*x^3 + E^(2*E^3)*(48*x^2 - 12*x^3) + E^E^3*(192*x - 96*x^2 + 12*x^3)),x]

[Out]

(5*x^2*Log[4]^2)/(4*(1 - E^E^3)^2) + (5*(9 - E^E^3)^2*Log[4]^2)/((1 - E^E^3)^4*(4 - (1 - E^E^3)*x)^2) - (5*(9
- E^E^3)*(17 - E^E^3)*Log[4]^2)/(2*(1 - E^E^3)^4*(4 - (1 - E^E^3)*x)) - (5*x*(4 + 12*E^(2*E^3) - 4*E^(3*E^3) -
9*Log[4]^2 - E^E^3*(12 - Log[4]^2)))/(4*(1 - E^E^3)^3)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1280+960 x-240 x^2+20 x^3-20 e^{3 e^3} x^3+e^{2 e^3} \left (-240 x^2+60 x^3\right )+\left (10 x+60 x^2+75 x^3-10 x^4\right ) \log ^2(4)+e^{e^3} \left (-960 x+480 x^2-60 x^3+\left (5 x^3+10 x^4\right ) \log ^2(4)\right )}{256-192 x+48 x^2+\left (-4+4 e^{3 e^3}\right ) x^3+e^{2 e^3} \left (48 x^2-12 x^3\right )+e^{e^3} \left (192 x-96 x^2+12 x^3\right )} \, dx\\ &=\int \frac {-1280+960 x-240 x^2+\left (20-20 e^{3 e^3}\right ) x^3+e^{2 e^3} \left (-240 x^2+60 x^3\right )+\left (10 x+60 x^2+75 x^3-10 x^4\right ) \log ^2(4)+e^{e^3} \left (-960 x+480 x^2-60 x^3+\left (5 x^3+10 x^4\right ) \log ^2(4)\right )}{256-192 x+48 x^2+\left (-4+4 e^{3 e^3}\right ) x^3+e^{2 e^3} \left (48 x^2-12 x^3\right )+e^{e^3} \left (192 x-96 x^2+12 x^3\right )} \, dx\\ &=\int \left (\frac {5 x \log ^2(4)}{2 \left (-1+e^{e^3}\right )^2}+\frac {10 \left (9-e^{e^3}\right )^2 \log ^2(4)}{\left (1-e^{e^3}\right )^3 \left (4-\left (1-e^{e^3}\right ) x\right )^3}+\frac {5 \left (9-e^{e^3}\right ) \left (-17+e^{e^3}\right ) \log ^2(4)}{2 \left (1-e^{e^3}\right )^3 \left (4-\left (1-e^{e^3}\right ) x\right )^2}+\frac {15 e^{e^3} \left (1+\frac {e^{-e^3} \left (4-4 e^{3 e^3}-9 \log ^2(4)+e^{e^3} \log ^2(4)\right )}{12 \left (-1+e^{e^3}\right )}\right )}{\left (-1+e^{e^3}\right )^2}\right ) \, dx\\ &=\frac {5 x^2 \log ^2(4)}{4 \left (1-e^{e^3}\right )^2}+\frac {5 \left (9-e^{e^3}\right )^2 \log ^2(4)}{\left (1-e^{e^3}\right )^4 \left (4-\left (1-e^{e^3}\right ) x\right )^2}-\frac {5 \left (9-e^{e^3}\right ) \left (17-e^{e^3}\right ) \log ^2(4)}{2 \left (1-e^{e^3}\right )^4 \left (4-\left (1-e^{e^3}\right ) x\right )}-\frac {5 x \left (4+12 e^{2 e^3}-4 e^{3 e^3}-9 \log ^2(4)-e^{e^3} \left (12-\log ^2(4)\right )\right )}{4 \left (1-e^{e^3}\right )^3}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.22, size = 131, normalized size = 3.54 \begin {gather*} \frac {5 \left (\left (-1+e^{e^3}\right )^2 x^2 \log ^2(4)-\frac {2 \left (-9+e^{e^3}\right ) \left (-50+e^{e^3} (2-18 x)+17 x+e^{2 e^3} x\right ) \log ^2(4)}{\left (4+\left (-1+e^{e^3}\right ) x\right )^2}-\left (-1+e^{e^3}\right ) x \left (-4-12 e^{2 e^3}+4 e^{3 e^3}+9 \log ^2(4)-e^{e^3} \left (-12+\log ^2(4)\right )\right )\right )}{4 \left (-1+e^{e^3}\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1280 + 960*x - 240*x^2 + 20*x^3 - 20*E^(3*E^3)*x^3 + E^(2*E^3)*(-240*x^2 + 60*x^3) + (10*x + 60*x^
2 + 75*x^3 - 10*x^4)*Log[4]^2 + E^E^3*(-960*x + 480*x^2 - 60*x^3 + (5*x^3 + 10*x^4)*Log[4]^2))/(256 - 192*x +
48*x^2 - 4*x^3 + 4*E^(3*E^3)*x^3 + E^(2*E^3)*(48*x^2 - 12*x^3) + E^E^3*(192*x - 96*x^2 + 12*x^3)),x]

[Out]

(5*((-1 + E^E^3)^2*x^2*Log[4]^2 - (2*(-9 + E^E^3)*(-50 + E^E^3*(2 - 18*x) + 17*x + E^(2*E^3)*x)*Log[4]^2)/(4 +
(-1 + E^E^3)*x)^2 - (-1 + E^E^3)*x*(-4 - 12*E^(2*E^3) + 4*E^(3*E^3) + 9*Log[4]^2 - E^E^3*(-12 + Log[4]^2))))/
(4*(-1 + E^E^3)^4)

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fricas [B]  time = 0.70, size = 330, normalized size = 8.92 \begin {gather*} -\frac {5 \, {\left (x^{3} e^{\left (6 \, e^{3}\right )} + x^{3} - {\left (x^{4} + x^{3} - 56 \, x^{2} + 450 \, x - 900\right )} \log \relax (2)^{2} - 8 \, x^{2} - 2 \, {\left (3 \, x^{3} - 4 \, x^{2}\right )} e^{\left (5 \, e^{3}\right )} + {\left (15 \, x^{3} - {\left (x^{4} + x^{3}\right )} \log \relax (2)^{2} - 40 \, x^{2} + 16 \, x\right )} e^{\left (4 \, e^{3}\right )} - 2 \, {\left (10 \, x^{3} - {\left (2 \, x^{4} + 2 \, x^{3} - 4 \, x^{2} + x\right )} \log \relax (2)^{2} - 40 \, x^{2} + 32 \, x\right )} e^{\left (3 \, e^{3}\right )} + {\left (15 \, x^{3} - 2 \, {\left (3 \, x^{4} + 3 \, x^{3} - 36 \, x^{2} + 35 \, x - 2\right )} \log \relax (2)^{2} - 80 \, x^{2} + 96 \, x\right )} e^{\left (2 \, e^{3}\right )} - 2 \, {\left (3 \, x^{3} - {\left (2 \, x^{4} + 2 \, x^{3} - 60 \, x^{2} + 259 \, x - 68\right )} \log \relax (2)^{2} - 20 \, x^{2} + 32 \, x\right )} e^{\left (e^{3}\right )} + 16 \, x\right )}}{x^{2} e^{\left (6 \, e^{3}\right )} + x^{2} - 2 \, {\left (3 \, x^{2} - 4 \, x\right )} e^{\left (5 \, e^{3}\right )} + {\left (15 \, x^{2} - 40 \, x + 16\right )} e^{\left (4 \, e^{3}\right )} - 4 \, {\left (5 \, x^{2} - 20 \, x + 16\right )} e^{\left (3 \, e^{3}\right )} + {\left (15 \, x^{2} - 80 \, x + 96\right )} e^{\left (2 \, e^{3}\right )} - 2 \, {\left (3 \, x^{2} - 20 \, x + 32\right )} e^{\left (e^{3}\right )} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x^3*exp(exp(3))^3+(60*x^3-240*x^2)*exp(exp(3))^2+(4*(10*x^4+5*x^3)*log(2)^2-60*x^3+480*x^2-960*
x)*exp(exp(3))+4*(-10*x^4+75*x^3+60*x^2+10*x)*log(2)^2+20*x^3-240*x^2+960*x-1280)/(4*x^3*exp(exp(3))^3+(-12*x^
3+48*x^2)*exp(exp(3))^2+(12*x^3-96*x^2+192*x)*exp(exp(3))-4*x^3+48*x^2-192*x+256),x, algorithm="fricas")

[Out]

-5*(x^3*e^(6*e^3) + x^3 - (x^4 + x^3 - 56*x^2 + 450*x - 900)*log(2)^2 - 8*x^2 - 2*(3*x^3 - 4*x^2)*e^(5*e^3) +
(15*x^3 - (x^4 + x^3)*log(2)^2 - 40*x^2 + 16*x)*e^(4*e^3) - 2*(10*x^3 - (2*x^4 + 2*x^3 - 4*x^2 + x)*log(2)^2 -
40*x^2 + 32*x)*e^(3*e^3) + (15*x^3 - 2*(3*x^4 + 3*x^3 - 36*x^2 + 35*x - 2)*log(2)^2 - 80*x^2 + 96*x)*e^(2*e^3
) - 2*(3*x^3 - (2*x^4 + 2*x^3 - 60*x^2 + 259*x - 68)*log(2)^2 - 20*x^2 + 32*x)*e^(e^3) + 16*x)/(x^2*e^(6*e^3)
+ x^2 - 2*(3*x^2 - 4*x)*e^(5*e^3) + (15*x^2 - 40*x + 16)*e^(4*e^3) - 4*(5*x^2 - 20*x + 16)*e^(3*e^3) + (15*x^2
- 80*x + 96)*e^(2*e^3) - 2*(3*x^2 - 20*x + 32)*e^(e^3) - 8*x + 16)

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giac [B]  time = 1.73, size = 334, normalized size = 9.03 \begin {gather*} \frac {5 \, {\left (x^{2} e^{\left (4 \, e^{3}\right )} \log \relax (2)^{2} - 4 \, x^{2} e^{\left (3 \, e^{3}\right )} \log \relax (2)^{2} + 6 \, x^{2} e^{\left (2 \, e^{3}\right )} \log \relax (2)^{2} - 4 \, x^{2} e^{\left (e^{3}\right )} \log \relax (2)^{2} + x^{2} \log \relax (2)^{2} + x e^{\left (4 \, e^{3}\right )} \log \relax (2)^{2} - 12 \, x e^{\left (3 \, e^{3}\right )} \log \relax (2)^{2} + 30 \, x e^{\left (2 \, e^{3}\right )} \log \relax (2)^{2} - 28 \, x e^{\left (e^{3}\right )} \log \relax (2)^{2} + 9 \, x \log \relax (2)^{2} - x e^{\left (6 \, e^{3}\right )} + 6 \, x e^{\left (5 \, e^{3}\right )} - 15 \, x e^{\left (4 \, e^{3}\right )} + 20 \, x e^{\left (3 \, e^{3}\right )} - 15 \, x e^{\left (2 \, e^{3}\right )} + 6 \, x e^{\left (e^{3}\right )} - x\right )}}{e^{\left (6 \, e^{3}\right )} - 6 \, e^{\left (5 \, e^{3}\right )} + 15 \, e^{\left (4 \, e^{3}\right )} - 20 \, e^{\left (3 \, e^{3}\right )} + 15 \, e^{\left (2 \, e^{3}\right )} - 6 \, e^{\left (e^{3}\right )} + 1} - \frac {5 \, {\left (3.68889104300000 \times 10^{121} \, \log \left (x + 7.56902073509000 \times 10^{-9}\right ) + 3.12914073993000 \times 10^{121} \, \log \left (x + 7.56838262681000 \times 10^{-9}\right ) + 4.56113883361000 \times 10^{119} \, \log \left (x + 7.56596396813000 \times 10^{-9}\right )\right )}}{3 \, {\left (e^{\left (15 \, e^{3}\right )} - 15 \, e^{\left (14 \, e^{3}\right )} + 105 \, e^{\left (13 \, e^{3}\right )} - 455 \, e^{\left (12 \, e^{3}\right )} + 1365 \, e^{\left (11 \, e^{3}\right )} - 3003 \, e^{\left (10 \, e^{3}\right )} + 5005 \, e^{\left (9 \, e^{3}\right )} - 6435 \, e^{\left (8 \, e^{3}\right )} + 6435 \, e^{\left (7 \, e^{3}\right )} - 5005 \, e^{\left (6 \, e^{3}\right )} + 3003 \, e^{\left (5 \, e^{3}\right )} - 1365 \, e^{\left (4 \, e^{3}\right )} + 455 \, e^{\left (3 \, e^{3}\right )} - 105 \, e^{\left (2 \, e^{3}\right )} + 15 \, e^{\left (e^{3}\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x^3*exp(exp(3))^3+(60*x^3-240*x^2)*exp(exp(3))^2+(4*(10*x^4+5*x^3)*log(2)^2-60*x^3+480*x^2-960*
x)*exp(exp(3))+4*(-10*x^4+75*x^3+60*x^2+10*x)*log(2)^2+20*x^3-240*x^2+960*x-1280)/(4*x^3*exp(exp(3))^3+(-12*x^
3+48*x^2)*exp(exp(3))^2+(12*x^3-96*x^2+192*x)*exp(exp(3))-4*x^3+48*x^2-192*x+256),x, algorithm="giac")

[Out]

5*(x^2*e^(4*e^3)*log(2)^2 - 4*x^2*e^(3*e^3)*log(2)^2 + 6*x^2*e^(2*e^3)*log(2)^2 - 4*x^2*e^(e^3)*log(2)^2 + x^2
*log(2)^2 + x*e^(4*e^3)*log(2)^2 - 12*x*e^(3*e^3)*log(2)^2 + 30*x*e^(2*e^3)*log(2)^2 - 28*x*e^(e^3)*log(2)^2 +
9*x*log(2)^2 - x*e^(6*e^3) + 6*x*e^(5*e^3) - 15*x*e^(4*e^3) + 20*x*e^(3*e^3) - 15*x*e^(2*e^3) + 6*x*e^(e^3) -
x)/(e^(6*e^3) - 6*e^(5*e^3) + 15*e^(4*e^3) - 20*e^(3*e^3) + 15*e^(2*e^3) - 6*e^(e^3) + 1) - 5/3*(3.6888910430
0000e121*log(x + 7.56902073509000e-9) + 3.12914073993000e121*log(x + 7.56838262681000e-9) + 4.56113883361000e1
19*log(x + 7.56596396813000e-9))/(e^(15*e^3) - 15*e^(14*e^3) + 105*e^(13*e^3) - 455*e^(12*e^3) + 1365*e^(11*e^
3) - 3003*e^(10*e^3) + 5005*e^(9*e^3) - 6435*e^(8*e^3) + 6435*e^(7*e^3) - 5005*e^(6*e^3) + 3003*e^(5*e^3) - 13
65*e^(4*e^3) + 455*e^(3*e^3) - 105*e^(2*e^3) + 15*e^(e^3) - 1)

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maple [B]  time = 0.22, size = 68, normalized size = 1.84

method	result	size
norman	\(\frac {-80 x +\left (-5 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}+5 \ln \relax (2)^{2}+10 \,{\mathrm e}^{{\mathrm e}^{3}}-5\right ) x^{3}+\left (\frac {5 \ln \relax (2)^{2}}{4}-40 \,{\mathrm e}^{{\mathrm e}^{3}}+40\right ) x^{2}+5 x^{4} \ln \relax (2)^{2}}{\left (x \,{\mathrm e}^{{\mathrm e}^{3}}-x +4\right )^{2}}\)	\(68\)
gosper	\(-\frac {5 x \left (-4 x^{3} \ln \relax (2)^{2}+4 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} x^{2}-4 x^{2} \ln \relax (2)^{2}-8 x^{2} {\mathrm e}^{{\mathrm e}^{3}}-x \ln \relax (2)^{2}+32 x \,{\mathrm e}^{{\mathrm e}^{3}}+4 x^{2}-32 x +64\right )}{4 \left ({\mathrm e}^{2 \,{\mathrm e}^{3}} x^{2}-2 x^{2} {\mathrm e}^{{\mathrm e}^{3}}+8 x \,{\mathrm e}^{{\mathrm e}^{3}}+x^{2}-8 x +16\right )}\)	\(96\)
risch	\(\frac {5 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} x^{2}}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}+\frac {5 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} x}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}-\frac {5 \ln \relax (2)^{2} x^{2}}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}-\frac {45 \ln \relax (2)^{2} x}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}-\frac {15 \,{\mathrm e}^{{\mathrm e}^{3}} x}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}-\frac {5 x \,{\mathrm e}^{3 \,{\mathrm e}^{3}}}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}+\frac {15 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}+\frac {5 x}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right )}+\frac {\left (-10 \ln \relax (2)^{2} {\mathrm e}^{2 \,{\mathrm e}^{3}}+260 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2}-1530 \ln \relax (2)^{2}\right ) x -\frac {20 \ln \relax (2)^{2} \left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-34 \,{\mathrm e}^{{\mathrm e}^{3}}+225\right )}{{\mathrm e}^{{\mathrm e}^{3}}-1}}{\left ({\mathrm e}^{2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{{\mathrm e}^{3}}+1\right ) \left ({\mathrm e}^{{\mathrm e}^{3}}-1\right ) \left ({\mathrm e}^{2 \,{\mathrm e}^{3}} x^{2}-2 x^{2} {\mathrm e}^{{\mathrm e}^{3}}+8 x \,{\mathrm e}^{{\mathrm e}^{3}}+x^{2}-8 x +16\right )}\)	\(343\)
default	\(\frac {30 \ln \relax (2)^{2} {\mathrm e}^{2 \,{\mathrm e}^{3}} x^{2}+5 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} {\mathrm e}^{3 \,{\mathrm e}^{3}} x^{2}-15 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} {\mathrm e}^{2 \,{\mathrm e}^{3}} x^{2}+150 \ln \relax (2)^{2} {\mathrm e}^{2 \,{\mathrm e}^{3}} x +5 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} {\mathrm e}^{3 \,{\mathrm e}^{3}} x -135 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} {\mathrm e}^{2 \,{\mathrm e}^{3}} x -20 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} x^{2}-5 \ln \relax (2)^{2} {\mathrm e}^{3 \,{\mathrm e}^{3}} x^{2}-140 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (2)^{2} x +75 \ln \relax (2)^{2} {\mathrm e}^{3 \,{\mathrm e}^{3}} x +5 x^{2} \ln \relax (2)^{2}-75 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}-30 \,{\mathrm e}^{{\mathrm e}^{3}} {\mathrm e}^{3 \,{\mathrm e}^{3}} x +90 \,{\mathrm e}^{{\mathrm e}^{3}} {\mathrm e}^{2 \,{\mathrm e}^{3}} x +45 x \ln \relax (2)^{2}-5 x \,{\mathrm e}^{6 \,{\mathrm e}^{3}}+30 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} {\mathrm e}^{3 \,{\mathrm e}^{3}} x -45 x \,{\mathrm e}^{4 \,{\mathrm e}^{3}}+30 x \,{\mathrm e}^{{\mathrm e}^{3}}+10 x \,{\mathrm e}^{3 \,{\mathrm e}^{3}}-5 x}{\left (3 \,{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{3 \,{\mathrm e}^{3}}-3 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}-1\right )^{2}}-\frac {10 \left (\munderset {\textit {\_R} =\RootOf \left (-64-\left (3 \,{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{3 \,{\mathrm e}^{3}}-3 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}-1\right ) \textit {\_Z}^{3}-\left (-24 \,{\mathrm e}^{{\mathrm e}^{3}}+12 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}+12\right ) \textit {\_Z}^{2}-\left (48 \,{\mathrm e}^{{\mathrm e}^{3}}-48\right ) \textit {\_Z} \right )}{\sum }\frac {\left (638 \textit {\_R} \,{\mathrm e}^{{\mathrm e}^{3}}+30 \textit {\_R} \,{\mathrm e}^{5 \,{\mathrm e}^{3}}-263 \textit {\_R} \,{\mathrm e}^{4 \,{\mathrm e}^{3}}-\textit {\_R} \,{\mathrm e}^{6 \,{\mathrm e}^{3}}+772 \textit {\_R} \,{\mathrm e}^{3 \,{\mathrm e}^{3}}-1023 \textit {\_R} \,{\mathrm e}^{2 \,{\mathrm e}^{3}}-896 \,{\mathrm e}^{{\mathrm e}^{3}}+32 \,{\mathrm e}^{4 \,{\mathrm e}^{3}}-384 \,{\mathrm e}^{3 \,{\mathrm e}^{3}}+960 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}-153 \textit {\_R} +288\right ) \ln \left (x -\textit {\_R} \right )}{-16+\textit {\_R}^{2} {\mathrm e}^{3 \,{\mathrm e}^{3}}-3 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \textit {\_R}^{2}+8 \textit {\_R} \,{\mathrm e}^{2 \,{\mathrm e}^{3}}+3 \textit {\_R}^{2} {\mathrm e}^{{\mathrm e}^{3}}-16 \textit {\_R} \,{\mathrm e}^{{\mathrm e}^{3}}-\textit {\_R}^{2}+16 \,{\mathrm e}^{{\mathrm e}^{3}}+8 \textit {\_R}}\right ) \ln \relax (2)^{2}}{3 \left (3 \,{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{3 \,{\mathrm e}^{3}}-3 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}-1\right )^{2}}\)	\(519\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*x^3*exp(exp(3))^3+(60*x^3-240*x^2)*exp(exp(3))^2+(4*(10*x^4+5*x^3)*ln(2)^2-60*x^3+480*x^2-960*x)*exp(
exp(3))+4*(-10*x^4+75*x^3+60*x^2+10*x)*ln(2)^2+20*x^3-240*x^2+960*x-1280)/(4*x^3*exp(exp(3))^3+(-12*x^3+48*x^2
)*exp(exp(3))^2+(12*x^3-96*x^2+192*x)*exp(exp(3))-4*x^3+48*x^2-192*x+256),x,method=_RETURNVERBOSE)

[Out]

(-80*x+(-5*exp(exp(3))^2+5*ln(2)^2+10*exp(exp(3))-5)*x^3+(5/4*ln(2)^2-40*exp(exp(3))+40)*x^2+5*x^4*ln(2)^2)/(x
*exp(exp(3))-x+4)^2

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maxima [B]  time = 0.44, size = 227, normalized size = 6.14 \begin {gather*} -\frac {10 \, {\left (x {\left (e^{\left (3 \, e^{3}\right )} - 27 \, e^{\left (2 \, e^{3}\right )} + 179 \, e^{\left (e^{3}\right )} - 153\right )} \log \relax (2)^{2} + 2 \, {\left (e^{\left (2 \, e^{3}\right )} - 34 \, e^{\left (e^{3}\right )} + 225\right )} \log \relax (2)^{2}\right )}}{x^{2} {\left (e^{\left (6 \, e^{3}\right )} - 6 \, e^{\left (5 \, e^{3}\right )} + 15 \, e^{\left (4 \, e^{3}\right )} - 20 \, e^{\left (3 \, e^{3}\right )} + 15 \, e^{\left (2 \, e^{3}\right )} - 6 \, e^{\left (e^{3}\right )} + 1\right )} + 8 \, x {\left (e^{\left (5 \, e^{3}\right )} - 5 \, e^{\left (4 \, e^{3}\right )} + 10 \, e^{\left (3 \, e^{3}\right )} - 10 \, e^{\left (2 \, e^{3}\right )} + 5 \, e^{\left (e^{3}\right )} - 1\right )} + 16 \, e^{\left (4 \, e^{3}\right )} - 64 \, e^{\left (3 \, e^{3}\right )} + 96 \, e^{\left (2 \, e^{3}\right )} - 64 \, e^{\left (e^{3}\right )} + 16} + \frac {5 \, {\left (x^{2} {\left (e^{\left (e^{3}\right )} - 1\right )} \log \relax (2)^{2} + {\left ({\left (e^{\left (e^{3}\right )} - 9\right )} \log \relax (2)^{2} - e^{\left (3 \, e^{3}\right )} + 3 \, e^{\left (2 \, e^{3}\right )} - 3 \, e^{\left (e^{3}\right )} + 1\right )} x\right )}}{e^{\left (3 \, e^{3}\right )} - 3 \, e^{\left (2 \, e^{3}\right )} + 3 \, e^{\left (e^{3}\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x^3*exp(exp(3))^3+(60*x^3-240*x^2)*exp(exp(3))^2+(4*(10*x^4+5*x^3)*log(2)^2-60*x^3+480*x^2-960*
x)*exp(exp(3))+4*(-10*x^4+75*x^3+60*x^2+10*x)*log(2)^2+20*x^3-240*x^2+960*x-1280)/(4*x^3*exp(exp(3))^3+(-12*x^
3+48*x^2)*exp(exp(3))^2+(12*x^3-96*x^2+192*x)*exp(exp(3))-4*x^3+48*x^2-192*x+256),x, algorithm="maxima")

[Out]

-10*(x*(e^(3*e^3) - 27*e^(2*e^3) + 179*e^(e^3) - 153)*log(2)^2 + 2*(e^(2*e^3) - 34*e^(e^3) + 225)*log(2)^2)/(x
^2*(e^(6*e^3) - 6*e^(5*e^3) + 15*e^(4*e^3) - 20*e^(3*e^3) + 15*e^(2*e^3) - 6*e^(e^3) + 1) + 8*x*(e^(5*e^3) - 5
*e^(4*e^3) + 10*e^(3*e^3) - 10*e^(2*e^3) + 5*e^(e^3) - 1) + 16*e^(4*e^3) - 64*e^(3*e^3) + 96*e^(2*e^3) - 64*e^
(e^3) + 16) + 5*(x^2*(e^(e^3) - 1)*log(2)^2 + ((e^(e^3) - 9)*log(2)^2 - e^(3*e^3) + 3*e^(2*e^3) - 3*e^(e^3) +
1)*x)/(e^(3*e^3) - 3*e^(2*e^3) + 3*e^(e^3) - 1)

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mupad [B]  time = 1.02, size = 236, normalized size = 6.38 \begin {gather*} \frac {x\,\left (10\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,{\ln \relax (2)}^2+1530\,{\ln \relax (2)}^2-260\,{\mathrm {e}}^{{\mathrm {e}}^3}\,{\ln \relax (2)}^2\right )+\frac {20\,\left ({\mathrm {e}}^{2\,{\mathrm {e}}^3}\,{\ln \relax (2)}^2+225\,{\ln \relax (2)}^2-34\,{\mathrm {e}}^{{\mathrm {e}}^3}\,{\ln \relax (2)}^2\right )}{{\mathrm {e}}^{{\mathrm {e}}^3}-1}}{\left (10\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}-10\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}+5\,{\mathrm {e}}^{4\,{\mathrm {e}}^3}-{\mathrm {e}}^{5\,{\mathrm {e}}^3}-5\,{\mathrm {e}}^{{\mathrm {e}}^3}+1\right )\,x^2+\left (32\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}-48\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}-8\,{\mathrm {e}}^{4\,{\mathrm {e}}^3}+32\,{\mathrm {e}}^{{\mathrm {e}}^3}-8\right )\,x+48\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}-16\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}-48\,{\mathrm {e}}^{{\mathrm {e}}^3}+16}+x\,\left (\frac {15\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}-5\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}-15\,{\mathrm {e}}^{{\mathrm {e}}^3}+75\,{\ln \relax (2)}^2+5\,{\mathrm {e}}^{{\mathrm {e}}^3}\,{\ln \relax (2)}^2+5}{{\left ({\mathrm {e}}^{{\mathrm {e}}^3}-1\right )}^3}-\frac {120\,{\ln \relax (2)}^2}{{\left ({\mathrm {e}}^{{\mathrm {e}}^3}-1\right )}^3}\right )+\frac {5\,x^2\,{\ln \relax (2)}^2}{{\left ({\mathrm {e}}^{{\mathrm {e}}^3}-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(3))*(240*x^2 - 60*x^3) - 4*log(2)^2*(10*x + 60*x^2 + 75*x^3 - 10*x^4) - 960*x + 20*x^3*exp(3*e
xp(3)) + exp(exp(3))*(960*x - 480*x^2 + 60*x^3 - 4*log(2)^2*(5*x^3 + 10*x^4)) + 240*x^2 - 20*x^3 + 1280)/(exp(
2*exp(3))*(48*x^2 - 12*x^3) - 192*x + 4*x^3*exp(3*exp(3)) + exp(exp(3))*(192*x - 96*x^2 + 12*x^3) + 48*x^2 - 4
*x^3 + 256),x)

[Out]

(x*(10*exp(2*exp(3))*log(2)^2 + 1530*log(2)^2 - 260*exp(exp(3))*log(2)^2) + (20*(exp(2*exp(3))*log(2)^2 + 225*
log(2)^2 - 34*exp(exp(3))*log(2)^2))/(exp(exp(3)) - 1))/(48*exp(2*exp(3)) - 16*exp(3*exp(3)) - 48*exp(exp(3))
+ x^2*(10*exp(2*exp(3)) - 10*exp(3*exp(3)) + 5*exp(4*exp(3)) - exp(5*exp(3)) - 5*exp(exp(3)) + 1) - x*(48*exp(
2*exp(3)) - 32*exp(3*exp(3)) + 8*exp(4*exp(3)) - 32*exp(exp(3)) + 8) + 16) + x*((15*exp(2*exp(3)) - 5*exp(3*ex
p(3)) - 15*exp(exp(3)) + 75*log(2)^2 + 5*exp(exp(3))*log(2)^2 + 5)/(exp(exp(3)) - 1)^3 - (120*log(2)^2)/(exp(e
xp(3)) - 1)^3) + (5*x^2*log(2)^2)/(exp(exp(3)) - 1)^2

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sympy [B]  time = 1.92, size = 423, normalized size = 11.43 \begin {gather*} \frac {5 x^{2} \log {\relax (2 )}^{2}}{- 2 e^{e^{3}} + 1 + e^{2 e^{3}}} - x \left (- \frac {15 e^{2 e^{3}}}{- 3 e^{2 e^{3}} - 1 + 3 e^{e^{3}} + e^{3 e^{3}}} - \frac {5 e^{e^{3}} \log {\relax (2 )}^{2}}{- 3 e^{2 e^{3}} - 1 + 3 e^{e^{3}} + e^{3 e^{3}}} - \frac {5}{- 3 e^{2 e^{3}} - 1 + 3 e^{e^{3}} + e^{3 e^{3}}} + \frac {45 \log {\relax (2 )}^{2}}{- 3 e^{2 e^{3}} - 1 + 3 e^{e^{3}} + e^{3 e^{3}}} + \frac {15 e^{e^{3}}}{- 3 e^{2 e^{3}} - 1 + 3 e^{e^{3}} + e^{3 e^{3}}} + \frac {5 e^{3 e^{3}}}{- 3 e^{2 e^{3}} - 1 + 3 e^{e^{3}} + e^{3 e^{3}}}\right ) - \frac {x \left (- 270 e^{2 e^{3}} \log {\relax (2 )}^{2} - 1530 \log {\relax (2 )}^{2} + 1790 e^{e^{3}} \log {\relax (2 )}^{2} + 10 e^{3 e^{3}} \log {\relax (2 )}^{2}\right ) - 680 e^{e^{3}} \log {\relax (2 )}^{2} + 4500 \log {\relax (2 )}^{2} + 20 e^{2 e^{3}} \log {\relax (2 )}^{2}}{x^{2} \left (- 6 e^{5 e^{3}} - 20 e^{3 e^{3}} - 6 e^{e^{3}} + 1 + 15 e^{2 e^{3}} + 15 e^{4 e^{3}} + e^{6 e^{3}}\right ) + x \left (- 40 e^{4 e^{3}} - 80 e^{2 e^{3}} - 8 + 40 e^{e^{3}} + 80 e^{3 e^{3}} + 8 e^{5 e^{3}}\right ) - 64 e^{3 e^{3}} - 64 e^{e^{3}} + 16 + 96 e^{2 e^{3}} + 16 e^{4 e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x**3*exp(exp(3))**3+(60*x**3-240*x**2)*exp(exp(3))**2+(4*(10*x**4+5*x**3)*ln(2)**2-60*x**3+480*
x**2-960*x)*exp(exp(3))+4*(-10*x**4+75*x**3+60*x**2+10*x)*ln(2)**2+20*x**3-240*x**2+960*x-1280)/(4*x**3*exp(ex
p(3))**3+(-12*x**3+48*x**2)*exp(exp(3))**2+(12*x**3-96*x**2+192*x)*exp(exp(3))-4*x**3+48*x**2-192*x+256),x)

[Out]

5*x**2*log(2)**2/(-2*exp(exp(3)) + 1 + exp(2*exp(3))) - x*(-15*exp(2*exp(3))/(-3*exp(2*exp(3)) - 1 + 3*exp(exp
(3)) + exp(3*exp(3))) - 5*exp(exp(3))*log(2)**2/(-3*exp(2*exp(3)) - 1 + 3*exp(exp(3)) + exp(3*exp(3))) - 5/(-3
*exp(2*exp(3)) - 1 + 3*exp(exp(3)) + exp(3*exp(3))) + 45*log(2)**2/(-3*exp(2*exp(3)) - 1 + 3*exp(exp(3)) + exp
(3*exp(3))) + 15*exp(exp(3))/(-3*exp(2*exp(3)) - 1 + 3*exp(exp(3)) + exp(3*exp(3))) + 5*exp(3*exp(3))/(-3*exp(
2*exp(3)) - 1 + 3*exp(exp(3)) + exp(3*exp(3)))) - (x*(-270*exp(2*exp(3))*log(2)**2 - 1530*log(2)**2 + 1790*exp
(exp(3))*log(2)**2 + 10*exp(3*exp(3))*log(2)**2) - 680*exp(exp(3))*log(2)**2 + 4500*log(2)**2 + 20*exp(2*exp(3
))*log(2)**2)/(x**2*(-6*exp(5*exp(3)) - 20*exp(3*exp(3)) - 6*exp(exp(3)) + 1 + 15*exp(2*exp(3)) + 15*exp(4*exp
(3)) + exp(6*exp(3))) + x*(-40*exp(4*exp(3)) - 80*exp(2*exp(3)) - 8 + 40*exp(exp(3)) + 80*exp(3*exp(3)) + 8*ex
p(5*exp(3))) - 64*exp(3*exp(3)) - 64*exp(exp(3)) + 16 + 96*exp(2*exp(3)) + 16*exp(4*exp(3)))

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